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4x^2-33x-46=0
a = 4; b = -33; c = -46;
Δ = b2-4ac
Δ = -332-4·4·(-46)
Δ = 1825
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-5\sqrt{73}}{2*4}=\frac{33-5\sqrt{73}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+5\sqrt{73}}{2*4}=\frac{33+5\sqrt{73}}{8} $
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